The Economic Policy Institute periodically issues reports on wages of entry-level workers. The institute reported that entry-level wages for male college graduates were $21.68 per hour and for female college graduates were $18.80 per hour in 2011. Assume that the standard deviation for male graduates is $2.30, and for female graduates it is $2.05.

What is the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68?

What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80?

What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?

a) P(21.18 ≤ x ≤ 22.18) = 0.87644

b) P(18.30 ≤ x ≤ 19.30) =0.91456

c) P(x > 18.50) = 0.9452

Step-by-step explanation:

The sample mean is related to the population mean through

μₓ = μ

For the male graduates,

μₓ = μ = $21.68

For female graduates,

μₓ = μ = $18.80

Standard deviation for the male graduates = $2.30

Standard deviation for the female graduates = $2.05

And the standard deviation of the distribution of sample means is related to the population standard deviation through

σₓ = σ/√n

where n = Sample size

a) probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68?

Standard deviation of the distribution of sample means = σₓ = (2.30/√50) = $0.325

sample means that are $0.5 of the population mean range from ($21.18 and $22.18) (that is, mean ± $0.5)

Required probability = P(21.18 ≤ x ≤ 22.18)

This is a normal distribution problem

We first normalize $21.18 and $22.18

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For $21.18,

z = (x - μ)/σ = (21.18 - 21.68)/0.325 = -1.54

For $22.18,

z = (x - μ)/σ = (22.18 - 21.68)/0.325 = 1.54

Required probability

P(21.18 ≤ x ≤ 22.18) = P(-1.54 ≤ z ≤ 1.54)

We'll use data from the normal probability table for these probabilities

P(21.18 ≤ x ≤ 22.18) = P(-1.54 ≤ z ≤ 1.54)

= P(z ≤ 1.54) - (P ≤ -1.54)

= 0.93822 - 0.06178

= 0.87644

b) What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80?

Standard deviation of the distribution of sample means = σₓ = (2.05/√50) = $0.290

sample means that are $0.5 of the population mean range from ($18.30 and $19.30)

Required probability = P(18.30 ≤ x ≤ 19.30)

This is a normal distribution problem

We first normalize $18.30 and $19.30

For $18.30

z = (x - μ)/σ = (18.30 - 18.80)/0.290 = -1.72

For $19.30,

z = (x - μ)/σ = (19.30 - 18.80)/0.290 = 1.72

Required probability

P(18.30 ≤ x ≤ 19.30) = P(-1.72 ≤ z ≤ 1.72)

We'll use data from the normal probability table for these probabilities

P(18.30 ≤ x ≤ 19.30) = P(-1.72 ≤ z ≤ 1.72)

= P(z ≤ 1.72) - (P ≤ -1.72)

= 0.95728 - 0.04272

= 0.91456

What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?

σₓ = (2.05/√120) = $0.187

sample means that are more $0.3 below the population mean range from ($18.50)

Required probability = P(x > 18.50)

This is a normal distribution problem

We first normalize $18.50

z = (x - μ)/σ = (18.50 - 18.80)/0.187 = -1.60

Required probability

P(x > 18.50) = P(z > -1.60)

We'll use data from the normal probability table for these probabilities

P(x > 18.50) = P(z > -1.60) = 1 - P(z ≤ -1.60)

= 1 - 0.05480 = 0.9452

Hope this Helps!!!